Calculate the sum of the geometric series $1+\left(\frac{1}{3}\right)+\left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^3 + \dots$. Express your answer as a common fraction.
This is an infinite geometric series with first term $a=1$ and common ratio $r=\frac{1}{3}$. Thus the sum is: $$\frac{a}{1-r} = \frac{1}{1-\frac13} = \frac{1}{\frac{2}{3}}=\boxed{\frac{3}{2}}.$$